Question

A particle is thrown vertically upward with initial velocity of 150m/s. Find the ratio of its speed at t=3s and t=5s. (Take g=10 \(ms^{-2}\))

Answer 1

Let's consider the upward direction as positive.
The initial velocity of the particle, u = 150 \(\frac{m}{s}\)
Acceleration due to gravity, g = -10 \(\frac{m}{s^2}\) (negative sign as it acts in the downward direction)
Using the kinematic equation, we can find the velocity of the particle at any time t as:
v = u - gt
At t = 3 s, the velocity of the particle is:
v₁ = u - gt 
= 150 - 10(3) 
= 120 \(\frac{m}{s}\)
At t = 5 s, the velocity of the particle is:
v₂ = u - gt
= 150 - 10(5) 
= 100 \(\frac{m}{s}\)

Therefore, the ratio of the speed of the particle at t=3s to t=5s is:
\(\frac{v_1}{v_2}\) = \(\frac{120}{100}\) = \(\frac{6}{5}\)
\(\therefore\) the ratio of the speed of the particle at t=3s to t=5s is 6:5. i.e 1.20
 

Answer 2

The correct answer is 1.2
\(\frac{v_3}{v_5}=\frac{u-g\times3}{u-g\times5}\)
\(=(\frac{150-30}{150-50})\)
\(=(\frac{120}{100}) = \frac{6}{5}\)
= 1.2